I've heard a couple of explanations but none of them quite make sense. One was that the German rocket engineers used metric and the American rocket engineers used the English system and seconds were the only thing they could agree on. Is that true?


3 Answers 3


It's even simpler than a German-American disagreement. It's use of ambiguous units.

The term "specific X" means the amount of X you can get from a unit mass of something. For instance, in batteries, specific energy means the total amount of energy you can get from one unit mass of battery. As described in the Wikipedia article, *specific impulse" is the total impulse you can get from a unit mass of propellants. Total impulse is just force times the duration of the force (time), which is equal to the momentum imparted to the object the force is applied to.

But in the early days in the US, propellant quantities were measured on scales calibrated in pounds, or more accurately, pounds-force: the force resulting from one Earth g acting on one pound-mass. The two are not equivalent! The Wikipedia article about English and US customary unit systems describes the consistent resolution of these units — but unfortunately these were not used in the early rocket labs, like Goddard's.

When you do it right, as in proper SI units, total impulse is in Newton-seconds, and mass is in kilograms. Dividing the former by the latter gives meters per second, a velocity: the rocket's exhaust velocity.

But when you use the "pound" as both mass and force and treat them as equivalent, total impulse is pound-seconds and mass is pounds. When you divide the two the "pounds" cancel and you're left with just seconds.

It turns out that this number with units of seconds indeed has a physical interpretation: If you burn one pound-mass of propellant in an engine that produces one pound-force of thrust, that is the duration, in seconds, the engine can operate before exhausting the one pound-mass propellant reservoir.

To get the exhaust velocity from that version of specific impulse you have to multiply by g, the true conversion constant between pounds-force and pounds-mass.

  • 14
    $\begingroup$ Note that the ambiguity between pound-mass and pound-force is an American confusion. From the SI perspective, you indeed have a mass in kilograms. But when you're launching from earth, the mass in kilograms linearly translates to a weight in Newton, and the conversion factor is given by Earth's gravity g. There's no confusion or ambiguity in SI units. $\endgroup$
    – MSalters
    Commented Jan 28, 2020 at 13:16
  • 4
    $\begingroup$ @MSalters according to Wikipedia anyway, German rocket engineers used to measure thrust in kilogram-force (en.wikipedia.org/wiki/Kilogram-force) -- so I guess they would have had the exact same ambiguity? $\endgroup$
    – ajd
    Commented Jan 28, 2020 at 13:33
  • 13
    $\begingroup$ @ajd: SI units are fairly new; the Germans had the MKS (Meter, Kilogram, Seconds) system in the 1940's. This was already metric, and as a direct precursor of SI it was pretty sane, but the newton was only standardized in 1946. Rather convenient for rocketry, as the German rocket engineers at that time were moving to the USA or USSR ;) $\endgroup$
    – MSalters
    Commented Jan 28, 2020 at 13:57
  • 5
    $\begingroup$ German engineers used, at that time, the 'Technisches Maßsystem'. The reference units were kilograms-force (kiloponds) for force, meters and seconds. Mass was a derived unit, the 'Technische Masseneinheit' or TME, defined as the mass that accelerated to 1 m/s2 when pushed with a force of one kilopond. $\endgroup$
    – xxavier
    Commented Jan 28, 2020 at 19:23
  • 4
    $\begingroup$ This would be a good answer if the paragraph beginning "It turns out that this number"... That physical interpretation has no value, and is never used. Well, never used by rocket scientists but used to explain the expression of seconds when discussing specific impulse. I suggest removing that misleading and useless paragraph, as it is only a distraction, and is not part of "the story" asked for in the title and OP. $\endgroup$
    – dotancohen
    Commented Jan 28, 2020 at 19:32

The term "specific" means "per unit." Specific impulse is defined as the impulse available per unit of fuel, and it has units of seconds when weight is chosen as the dividing unit. See this derivation by NASA for a step-by-step derivation of $I_{sp}$. I'll cover the important steps in this answer without relying on a unit system.

We begin with the equation for impulse: $$I = m V_{eq}$$

Dividing by weight: $$ \frac{I}{W} = \frac{m V_{eq}}{W} = I_{sp}$$

By Newton's second law, $F = ma \Rightarrow W = mg$

Substituting $mg$ for $W$: $$I_{sp} = \frac{m V_{eq}}{mg} = \frac{V_{eq}}{g}$$

As per the NASA derivation, $V_{eq} = \frac{F}{\dot{m}}$. Making this substitution: $$I_{sp} = \frac{\frac{F}{\dot{m}}}{g} = \frac{F}{\dot{m} g}$$

Now, let's do dimensional analysis: $$\left[I_{sp}\right] = \frac{\left[force\right]}{\frac{\left[mass\right]}{\left[time\right]} \frac{\left[length\right]}{\left[time\right]^2}}$$

Again by Newton's second law, we have $\left[force\right] = \left[mass\right] \frac{\left[length\right]}{\left[time\right]^2}$

The right-hand-side expression is found in the denominator of the previous equation. We have, finally, $$\left[I_{sp}\right] = \frac{\left[force\right]}{\frac{\left[force\right]}{\left[time\right]}} = \left[time\right]$$

So, both in US Customary units and SI units, $I_{sp}$ has units of seconds when weight is used as the dividing unit. It has nothing to do with agreeing upon units to use or "doing it right" and using "proper units."

  • $\begingroup$ The question here asks why "weight is chosen as the dividing unit". This answer doesn't really answer that question. $\endgroup$
    – JiK
    Commented Jan 29, 2020 at 21:19
  • $\begingroup$ @JiK Where do you see that being asked? I just reread the three sentences of the question several times and don't see how it's asking why weight is the dividing unit. If I had to guess I'd say it's because rocket fuel is measured by weight in practical applications (I don't know for a fact that this is true, but it seems reasonable given that weight is a key constraint in rocket design). Keeping as many calculations as possible in terms of weight would then reduce the likelihood of an error caused by someone using fuel mass where he should have used fuel weight. $\endgroup$
    – zaen
    Commented Jan 29, 2020 at 21:55
  • $\begingroup$ The term “specific” does not mean “per unit”, which wouldn’t make any sense. Actually, it means “divided by mass”. See for example ISO 80000-1: “The adjective “specific” is added to the name of a quantity to indicate the quotient of that quantity by mass.” or IUPAC: “The adjective specific before the name of an extensive quantity is used to mean divided by mass. $\endgroup$
    – user10840
    Commented Jan 30, 2020 at 17:56
  • $\begingroup$ @Loong ISO 80000-1 was published in 2009 and the Green Book (where your IUPAC quote comes from) was first published in 1988. Specific impulse was being used for quite a while before either of those. Regardless, if you're dead set on defining it as being divided by mass, then 1. pounds force and pounds mass don't cancel like the accepted answer states (you instead have pound-seconds per slug (lb*sec/sl) or pounds force * seconds per pound mass (lb$_f$*sec/lb$_m$)) and 2. you can't express $I_{sp}$ in seconds because that's "wrong," even though that's what the industry currently does. $\endgroup$
    – zaen
    Commented Jan 31, 2020 at 1:56
  • 1
    $\begingroup$ Also, see Wikipedia's list of specific quantities. This list is probably not exhaustive, but there are 18 divided by mass and 26 divided by other quantities, give or take a few since they have alternate definitions that include being divided by mass. I'm not even saying that most specific quantities are divided by units other than mass, but there are several significant ones (e.g. specific fuel consumption) that are divided by other units. $\endgroup$
    – zaen
    Commented Jan 31, 2020 at 2:22

The specific impulse $I_{\mathrm{sp}}$ in time units has the following interpretation. By burning propellant equal to a small fraction $\alpha$ of the rocket's total mass, the rocket can hover stationary in the air (in Earth gravity) for a short time $\tau$. The specific impulse is the proportionality factor: $\tau = \alpha I_{\mathrm{sp}}$. By integration, the rocket's total mass decreases with elapsed time $t$ as $\exp(-t/I_{\mathrm{sp}})$. That is, $I_{\mathrm{sp}}$ is the time constant for the decay of the rocket's mass when hovering.

And, since accelerating the rocket upward can only burn propellant faster, $I_{\mathrm{sp}}$ in time units sets a rough upper bound on how long a rocket can be fired during a launch from Earth. For example, the Space Shuttle's solid rocket boosters had a specific impulse of ~4 minutes and were fired for ~2 minutes.

  • 1
    $\begingroup$ " Isp in time units sets a rough upper bound on how long a rocket can be fired during a launch from Earth" - what about the engines designed to start in an orbit and continue to work in vacuum? How this definition applies to them? $\endgroup$ Commented Jan 29, 2020 at 10:04
  • 3
    $\begingroup$ I have never heard anyone express Isp in "minutes". And since your example of firing time has 100% error, it seems a poor usage. Do you have any reference for this unusual use of Isp? $\endgroup$ Commented Jan 29, 2020 at 14:40
  • 2
    $\begingroup$ @LeoS Such an engine's $I_{\mathrm{sp}}$ describes how long it could be effectively fired if it were used for vertical thrust during a launch. The same would apply if it is used for a subsequent ~1g "heavy lifting" maneuver such as going quickly from orbit to escape. But, if it is used in orbit to generate thrust much less than the total weight of the craft, then it could last much longer than $I_{\mathrm{sp}}$. All this is based on the idea that the available propellant is an order-one fraction of the craft's mass. $\endgroup$
    – nanoman
    Commented Jan 29, 2020 at 19:02
  • 1
    $\begingroup$ @OrganicMarble Well, if you can express $I_{\mathrm{sp}}$ in seconds, I can divide by 60 and express it in minutes. It's just normal unit manipulation. $\endgroup$
    – nanoman
    Commented Jan 29, 2020 at 19:04
  • 1
    $\begingroup$ @OrganicMarble I said $I_{\mathrm{sp}}$ "sets a rough upper bound", so I don't think there's a "100% error" in the example. Of course the rocket can be fired for a shorter time. My point is that $I_{\mathrm{sp}}$ in time units really does mean something in terms of time on a stopwatch. It's not a coincidence that for launches that take minutes, we need rockets with $I_{\mathrm{sp}}$ of minutes. ... $\endgroup$
    – nanoman
    Commented Jan 29, 2020 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.